The Riemann bilinear relations and the period lattice

If $g=0$, [F1] gives $H_1(X;\mathbb{Z})=0$, [F2] gives $\mathrm{\Omega }(X)=0$, and [F3] gives the zero period pairing; the normalized basis, symmetry, and strict-positivity statements are empty, while $e$ is the unique map $0\to 0$ and the rank-zero lattice convention in the Statement gives the point quotient. Thus all claims hold in this case, with positive definiteness vacuous on the zero vector space. For the rest of the proof assume $g\ge 1$.

Define $\mathcal{A}(\omega )=(P(a_i,\omega ){)}_{i=1}^g$. It is complex-linear by [F3]. If $\mathcal{A}(\omega )=0$, then [F4] gives ${\mathrm{\Pi }}_{\omega }(a_i)={\mathrm{\Pi }}_{\bar{\omega }}(a_i)=0$ for every $i$, so every summand of $S(\omega ,\bar{\omega })$ vanishes. By [F5], ${\int }_X\omega \wedge \bar{\omega }=0$. But [F6] makes $i\omega \wedge \bar{\omega }$ a nonnegative top form, and if $\omega \mathrm{\ne }0$ it is nonzero at a point; since $X$ is compact it is compactly supported, so [F7] gives $i{\int }_X\omega \wedge \bar{\omega }>0$, a contradiction. Thus $\mathcal{A}$ is injective.

By [F2], both domain and codomain of $\mathcal{A}$ have dimension $g$. Choose a basis ${\eta }_1,\dots ,{\eta }_g$ of $\mathrm{\Omega }(X)$; the matrix $M_{ij}=P(a_i,{\eta }_j)$ represents $\mathcal{A}$ and has zero kernel by step 2.1, so [F8] makes it invertible and $\mathcal{A}$ an isomorphism. Define ${\omega }_j={\mathcal{A}}^{-1}(e_j)$, where $e_j$ is the $j$th standard basis vector of ${\mathbb{C}}^g$. Then $P(a_i,{\omega }_j)={\delta }_{ij}$, and the isomorphism makes this normalized basis unique.

For holomorphic $\omega ,\eta$, [F6] gives $\omega \wedge \eta =0$, hence $S(\omega ,\eta )=0$ by [F5]. Write $\omega ={\sum }_j{c}_j{\omega }_j$ and $\eta ={\sum }_j{d}_j{\omega }_j$. Their $a$-period vectors are $c,d$ and their $b$-period vectors are $\mathrm{\Pi }c,\mathrm{\Pi }d$, so [F3] gives $S(\omega ,\eta )=c^{\mathsf{T}}(\mathrm{\Pi }-{\mathrm{\Pi }}^{\mathsf{T}})d$. This proves $S$ vanishes for all such pairs exactly when $\mathrm{\Pi }={\mathrm{\Pi }}^{\mathsf{T}}$: one direction follows from the displayed identity, and the reverse follows by setting $c=e_j,d=e_k$.

Put $Y=\mathrm{Im}\mathrm{\Pi }$, which is real symmetric by step 4.1. For $\omega ={\sum }_j{c}_j{\omega }_j$ with $c=x+iy$ and real $x,y$, [F4] and the definition of $\mathrm{\Pi }$ give $S(\omega ,\bar{\omega })=c^{\mathsf{T}}\bar{\mathrm{\Pi }}\bar{c}-(\mathrm{\Pi }c{)}^{\mathsf{T}}\bar{c}=c^{\mathsf{T}}(\bar{\mathrm{\Pi }}-\mathrm{\Pi })\bar{c}=-2i{c}^{\mathsf{T}}Y\bar{c}$, where symmetry of $\mathrm{\Pi }$ is used in the second equality. Since $Y$ is real symmetric, $c^{\mathsf{T}}Y\bar{c}=x^{\mathsf{T}}Yx+y^{\mathsf{T}}Yy$, and therefore $iS(\omega ,\bar{\omega })=2(x^{\mathsf{T}}Yx+y^{\mathsf{T}}Yy)$. By [F5]–[F7], the left side is strictly positive whenever $\omega \mathrm{\ne }0$. As $\omega \mapsto c$ is an isomorphism, this identity proves both directions of the equivalence: $Y$ is positive definite exactly when $iS(\omega ,\bar{\omega })>0$ for every nonzero $\omega$.

Write $\gamma ={\sum }_i(m_i{a}_i+n_i{b}_i)$ with $m,n\in {\mathbb{Z}}^g$, and identify a functional $\xi \in \mathrm{\Omega }(X{)}^{\ast }$ with $(\xi ({\omega }_1),\dots ,\xi ({\omega }_g))\in {\mathbb{C}}^g$. Its period vector is $(P(\gamma ,{\omega }_j){)}_{j=1}^g=m+{\mathrm{\Pi }}^{\mathsf{T}}n=m+\mathrm{\Pi }n$, by [F3] and symmetry. If $e(\gamma )=0$, taking imaginary parts gives $Yn=0$, hence $n=0$ and then $m=0$ by positivity of $Y$ from step 5.1; so $e$ is injective. For real $x,y\in {\mathbb{R}}^g$, a relation among the period vectors of the basis cycles has coordinates $x+{\mathrm{\Pi }}^{\mathsf{T}}y=0$; its imaginary part $Yy=0$ gives $y=0$ and then $x=0$. Thus those $2g$ vectors are real-linearly independent in ${\mathbb{C}}^g\cong {\mathbb{R}}^{2g}$, form a real basis, and by [F9] generate a full lattice. The displayed coordinate formula gives $\mathrm{\Lambda }={\mathbb{Z}}^g+\mathrm{\Pi }{\mathbb{Z}}^g$. The real-linear map $(x,y)\mapsto x+\mathrm{\Pi }y$ identifies ${\mathbb{R}}^{2g}\mathrm{/}{\mathbb{Z}}^{2g}$ with ${\mathbb{C}}^g\mathrm{/}\mathrm{\Lambda }$; the image of the compact cube $[0,1{]}^{2g}$ covers this quotient, so it is compact.